Getting to Mars: The Hohmann Transfer
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- Опубликовано: 12 сен 2024
- How long does it take to get to Mars? What Delta-Vs are required? When should you launch and why is a one way trip easier than a return mission? Mars One Astronaut Candidate Ryan MacDonald explains the Hohmann Transfer orbit.
*Part 1 (Orbital Mechanics 101): • Orbital Mechanics 101
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Thank you, this is so helpful! Me and my friend are doing a drama project about going to Mars and we wanted to make it as scientifically accurate as possible.
+Ruby Taylor
Glad to help!
Ok, I have to tell you, these 2 videos have genuinely helped me a lot, no other video I've found goes enough on the actual math to give me an understanding of where these numbers come from.
Whole heartedly thank you :
best video about orbital mechanics on youtube. everything is crystal clear and well explained. even 9 years after uploading this is still an exceptional video. thx bro
Came here from Kerbal Space Program :) Thanks for the great explanation!
You know strangly, when I watch your videos while I'm working (3D design/animation) it really keeps my mind clear and focused. And it's always good to hear your thoughts and explanations about mars one project. Whole globe is exited for it.
Thanks, I used to dabble in flash animation when I was younger (that's how I draw the 'Mission Update' graphics for my videos). It's always satisfying seeing something that was just an idea in your mind come to life :)
Actually, I was fascinated by your channel. So I was thinking making you a short opening animation, but I wasn't sure you might want it or not. If you want it, I would do it, but i want your opinion on that.
That sounds like a fantastic idea, I started work on one a few months ago but had to abandon it since .swf files aren't supported by the program I make my videos in.
Martian Colonist great! I think I have some ideas. Do you think we can move this talk to hangout?
jay choi Sure, I'm a little occupied at the moment sending off PhD applications though, so would you be available tomorrow or later in the week?
You broke the teachers law of mathematics:
Doing math in red pen
I was taught to double underline answers in red pen at university 😉
Glad he did that ..it went straight to head
this is really a awesome video.. such a easy and good explanation.. thank you sir for this video.
Wait wait wait... may I propose/ask for a follow-up to this one? It just ended abruptly and I was like "Wait, what? noooo... it's over I'm lost come back!"
This was great and it's helping me (finally) understand what we all just assume and never talk about (orbital mechanics). The math(s) isn't that painful which is just great.
Some ideas for a follow-up vid: the minutiae real rocket scientists have to deal with like not-perfect-circular orbits, Earth's atmosphere, mid-course corrections (how do they know where they are), mass issues, Lagrangian points, Buzz Aldrin's 'Mars Cycler" concept (would that make getting to/from mars easier/more efficient)- stuff like that from an orbital mechanics perspective. Just some ideas - I'm so glad I found your videos!
+Pete Kuhns
Hi Pete, thanks for your feedback!
I've had a plan for a while to make a 2-part video in which I derive Kepler's laws of motion from first principles (Newton's law of gravity), and hence show why planets orbit in ellipses. Basic rocket physics would also be a natural follow on, though I would have to increase the level of the Maths to include differential equations to explain this properly.
In the meantime, I'd be happy to Skype you to offer you some further assistance and ask for your ideas on future videos!
Thanks,
Ryan
+Pete Kuhns
Just to let you know that I have put together an extended video on Kepler's 1st law of planetary motion: ruclips.net/video/DurLVHPc1Iw/видео.html
+Martian Colonist Hi thank you! Watching now (safe for work right?) :-)
Thank you, not only did this video give me lots of new stuff to work on, but I also had a great lesson in rearranging formulae especially when finding Delta V1 and Delta V2.
It took two whole days to go through both vids in the series. Have definitely learned a lot.
Not the hero we wanted, but the hero we all needed
Thank you
I really appreciate you doing these videos Ryan! I am in no way as knowledgeable in physics and math as you, but I do hope to join you one day on Mars. I am trying to learn as much as I can before the fateful 'M-Day'. I plan to apply when applications reopen, but for now, these videos of yours help keep me informed about just what I'm getting myself into :) . Once again, thank you very much for providing a plethora of information in a way that even a novice of math like me can understand.
~I'll See you starside, Tyler Neumann
Great to meet you Tyler, and best of luck with your own application to the programme! I'll keep you updated when I have a more precise date for when the application process reopens.
Martian Colonist Glad to hear it! I'll make sure to keep a look out for your future videos :)
You really help me out with my homework. And those interesting real-life knowledge too! Thanks a lot! :)
You're awesome man! I'd been trying to solve some of these problems only using circular motion, but little did I know that circular motion is for noobs. Elliptical motion, the physics of the gods, was required. Evidently, you're now a god.
Sorry, it's the middle of the night here >.< Anyways, thanks again!
+Neoking
Just wait for my next video (this weekend), where I will derive elliptical orbits from first principles! Will then use that to derive Kepler's laws.
Circular motion is for noobs. 😆😆😆😆😆funny. But if the eccentricity is really small, circular motion should be accurate(oh wait, it's hohmann tranfer, never mind, noobs don't work here)
your videos are going to be the death of me if my teacher keeps giving them along with a heavy writing
assignment
I appologise on behalf of your teacher.
Very well done examples, explanations thru equating physical problems btw our Sun, Earth and Mars.
Thank you for the clear explanation! Good job :)
Very nice derivations of the deltaV values.
I know I am here 5 years late but I have a question. When you were substituting the equation from the angular momentum into the law of conservation equation why did you take out the mass of the rocket?
The mass of the rocket appears in both terms, and as we are seeking a relation for delta_v1 in terms of the other quantities it just cancels anyway.
It cancels out
The concepts here are so simple Newton's Law of Universal Gravitaion, Kepler's Law of Harmonies(Period), conservation of angular momentum and energy and angular velocity. But the way you apply it is even more complex and useful than the AP Physics 1 exam(college credit). These equations are also useful in docking(obviously attraction between spacecrafts are significant).
I meant arn't
Hi Ryan, thank you so much for your lessons! Now my work is more clear ;)
Elegant explanation. thank you very much for this informative video.
That trick to find angle theta is brilliant!
Thanks it helped lot in my exam
I'm doing this as a coding project and all of this is useful
Awesome! Thank you for sharing this. I just got stuck with the algebra when you wrote dv2 at 16:00, I know you took a common factor when doing dv1 and the rest is easy but in this case, i'm not sure if I see the subscripts right or as I suspect is my lack math skills... how sqrtGM/r2 - r1/r2 * sqrt2GMr2/(r1+r2)*r1 comes to be what you finally wrote as dv2?I know it is me struggling with algebra as usual but can't help trying to figure it out so I can move on to the next.Great channel, you have got a new subscriber!
+Jorge Calderon
Glad you liked the video!
There are a few more lines of algebra to get dv2, I've just written them down for you here: www.dropbox.com/s/4r6mtqx5m90w4ft/dv2%20derivaition.PNG?dl=0
Thanks,
Ryan
+Martian Colonist Wow! that was very nice, thank you so much, really appreciate it.
...and when you reach the surface you will need to use r2.d2*
@@martiancolonist Any chance you could make this available again? The link is broken, but I'm stuck on the same point!
I loved your final comment about what you would like your PhD to be in! How did you get on?
I finished a PhD in astronomy in 2019, with my research on exoplanet atmospheres. I'm now in the US as a Postdoctoral Research Associate in the same field.
@@martiancolonist Huge W
In the first part of the video, you calculated Δv assuming that the mass of the rocket stays constant.
Then for the second part you use the result of the Δv in the rocket equation, which is explicitly about an object changing mass. Could you explain this please?
Excellent video by the way, nice speed and structure.
Great question!
In the first part, this approximattion is essentially saying that the change in mass when executing an interplanetary transfer is negligable compared to that required to launch from a planetary surface. The calculation at the end with the rocket equation is just to illustrate the exponential dependence of the mass ratio on delta v.
@@martiancolonist Thanks! You're doing God's work.
This video is great helped me understand a lot :) I just wish this video was available in my native language then it would be even easier to understand :D
Whats your native language?
You'll get plenty of views from students coming from Tec de Monterrey
Nice simplification of the maths, though not simple enough for people who struggle with algebra. It's also ignoring some other big factors. ie it's assuming that Earth and Mars have zero mass and no atmosphere, hence no escape ΔV, no cost of climbing out of atmosphere, no capture gravity/atmo braking when arriving. Lunar gravity can also be a hurdle or a boon esp for orbital insertion vs atmo braking.
First of all, *great job on this presentation.*
If you don't mind, I have a couple of questions:
[1] At 1:12 you say that Mars is travelling in a _"circular orbit around the sun,"_ is this condition really necessary to solve this problem? I mean, just knowing r2 would not be enough?
[2] Do these calculations account for the variation in the rocket's mass as fuel is consumed on the way over to Mars?
[3] What about the effects of Mars's gravity on the rocket?
Velocity depends on the mass ratio. Escape velocity is a mass ratio atmosphere to that of the mass of the planet.
This is really well explained!
22:45 - That's exactly why there's so much interest in manufacturing methane fuel on Mars from local resources.
amazing they were able to make these calculations with the computers they had in the 1970s
Mr McDonald, this was a great presentation. Simply and elegantly explained. But is it that we ultimately solved the extra radiation problems of deep space (beyond our magnetic shield belts) for human exposure that well that we can put human on Mars?
If we have, wouldn’t it be nice to use the moon as an intermediary launch pad since the moon is moving faster than the earth?
And while on the moon, we don’t have to worry about falling out of orbit, so we could leave equipment there safely. Also, no atmosphere to destroy the craft during entries.
And we have mastered the man moon missions, so this should be viable.
So I'm guessing that what you mean by Years is the Period of the initial orbit right? It doesn't have to be earth years, it's just earth years because that's where we usually start from. Or is it Earth years because we're also using the Astronomical Unit.
Then if we were to start from Mars, p²=a³ where p is the period of mars and a is the SemiMajor axis as measured in distance from Mars right?
The use of Earth years is just to simplify Kepler's 3rd law. As you rightly noticed, it is when we use Earth years and astronomical units that P^2 = a^3. (there's a constant of proportionality that is one by definition when using years and AU, but is not one if you use other units).
Wouldnt the period depend on the gravitational force of the object u are flying around? So it cant depend on 'a' only. If a stays the same but gravity is increased, the object would have to fly faster and therefore have a smaller period. Even the units of a^3 (m^3) = p^2 (s^2) dont fit
An excellent observation! The equation a^3 = P^2 is actually only true when a is in AU, P is in years and you are orbiting the Sun. This holds because the AU is defined as the semi major axis for which the period is 1 year. In terms of the units, the equation is a^3 = k P^2, where the constant k = 1 AU^3 Yr^(-2), and so the constant is usually not written.
However, if you want the equation in SI units, it is: P^2 = (4*pi^2 / G*M) a^3. Here M is the mass of the object you are orbiting (here, the Sun). So you are absolutely right that the mass of the central object is important!
Where do you get the 7.4 from at the end? i feel like I have tried everything. But I am simply struggling to see the numbers I have to use in the equation.
Amazing video though, helped a lot!
2 questions:
- Did you pronounce Hohmann with an "f" as in Hofman? It's a german name, no 'f" sound in Hohmann
- How are you holding that pen? Where you taught that?
I think if you blast a automated pre fueled stage into earth orbit as cargo then dock up with it at the front you could blast off from orbit fully fueled up and could increase your speed and use the gravity assist to get to Mars much earlier .
😳😭 I would love to understand this...... Thanks for explaining it....
Out of curiosity, you mentioned that in real life we would have to deal with the inclination of Mars. However, wouldn't this be as simple as entering Low Earth orbit, then correcting the angle to match Mars? After this point you could essentially do everything else you talked about without worrying about inclination. Alternatively, could you simply launch at the modified angle? (sorry if this is a stupid question, I'm only in 9th grade so I have limited physics knowledge)
Also, forgot to mention. Awesome video. I learned a lot.
Really great video, great explanation! Thanks
It was a wonderful video! It helped a lot in my project but I had a small doubt(though I am here after 7 years), how is that you said the time period can vary from 6-8 months, isn't that variation too much? Despite the reasons you had stated, is there any mathematical proof that can justify your statement of 6-8 months ? I actually need this information for my project, would appreciate if you could reply.
Great question. The variation in the transfer time period isn't easy to show analytically on paper, since you have a multibody problem. This figure comes out of numerical calculations including the gravitational influence of all the planets, eccentric orbits, etc.
@@martiancolonist Hello. When you calculate the period of the Hohman transfrer orbit, why do you not include pi and standard gravitational perimeter? if you change the time to years and distance to Astronomical Units, does that value just cancel out? so would that be k that you mentioned?
Excellent. Thank you for the lesson.
Sir you broker the laws of physics
Sir when ever inserting the values in delta I am getting different answers so iam kindly requesting to make a video over this calculation
Sir iam participating in nss contest this calculation helps a lot in location transfer
Thank you very much for your nice and fàntastic video .
This was an amazing video keep up the good work
My compliments excellent Math lesson and explanation of the current targeting of Mars from Earth. I do not understand the reason for the elliptical transfer orbit approach to Mars from behind. This requires launching from the exterior of the earth/sun orbit to take advantage of the slinging effect of Earth. You can still get the slinging effect to the interior of the earth/sun orbit but head out the opposite way so that you are on collision course with Mars. Then you use some fuel to slow down and drive onto the surface. This should reduce the trip time by 66% No?
The Hohmann transfer is not the most time efficient transfer orbit, but it is the most energy efficient transfer (and hence uses the least fuel). By approaching Mars orbit head on, the amount by which you would have to change your speed to land on the surface would be twice the orbital velocity of Mars (2*24 km/s), which is almost 5 times the change in velocity needed to launch from Earth itself. Would be an extremely costly route to take!
I much appreciate your willingness to explain to my elementary understanding. When you refer to "changing speed to land" and "orbital velocity" does that take into account the potential for gravitational orbital capture by Mars and then slowing down more slowly?
Um, your correct if your just going from Earth orbit to Mars orbit on a Heliocentric elliptical Transfer but if your actually going from Earth to Mars you have to factor in deceleration as you leave Earth and acceleration as you approach Mars.
Need to calculate your hyperbolic exit velocity as you leave low earth orbit (LEO) such that it has the correct velocity when you transition from Earths SOI to the Suns SOI. DeltaV1 = hyperbolic exit velocity - orbital velocity at LEO.
Likewise when you arrive at Mars SOI you need to calculate the hyperbolic velocity at Periapsis based on having velocity Va (you calculated this) when you encounter Mars SOI. Delta V2 = hyperbolic velocity - orbital velocity at LMO.
If your transfer is along the orbital planes of Earth and Mars you will also need a minor inclination change as you change planes OR if you go ballisticaly you will have to adjust your inclination at either end.
Also rather than a circular capture you could go for a partial elliptical capture.
sorry
Yes, I am aware of all this. This video was intended as an introduction demonstrating the utility of constants of motion in the context of the Hohmann transfer. I stated the assumption of beginning in a 'high parking orbit' towards the beginning.
You must always be clear of the assumptions you are making but, at some point, you do have to settle on an agreed approximation. For example, if I'd decided to treat this using full General Relativity the problem would only be solvable numerically, which would defeat the purpose of illustrating how conservation of energy and momentum applies here algebraically :-)
cool 8-)
ps really like you other video on derivation or Kepler laws, hanging out for the next one
Kepler's Laws Part II is on the backburner at the moment (whilst I work on a video on Proxima b), but it will definitely be coming!
steamcommunity.com/sharedfiles/filedetails/?id=578501464
Hello! I just have a tiny question... In this video you say that Kepler's third law is P^2=a^3... In my schoolbooks it says P^2/a^3... How do you get the third law to be P^2=a^3?
+DasiyBrownCat
Kepler's third law in SI units can be written as: P^2 = (4pi^2/GM) a^3, which can be written as P^2/a^3 = (4pi^2/GM). If you choose units such that a is expressed in astronomical units and P in years, then the constant of proportionality is precisely 1 and Kepler's third law can be written as P^2/a^3 = 1.
I've got a dedicated video on Kepler's laws coming out this weekend, keep an eye out for it! :)
Love the video. Great work matey :)
At the beginning of the video, during the Hohmann transfer rundown, it's stated that the closest point is the perigee and farthest point the apogee. Technically, this isn't correct as the body being orbited is the sun; therefore, the correct terms would be perihelion and aphelion respectively. It's easy to accidentally use the wrong term when working with multiple bodies, which is why I prefer the generic apoapsis and periapsis terms.
You are indeed correct. An annotation was added to the video shortly after its release making this point, which perhaps you can't see on the device you watched the video on?
Martian Colonist ah yes, I'm on mobile so that's probably why. Thanks for replying. I just added the comment in case someone watching the video got confused about terminology. Great set of videos. Hope to see more.
the Hohmann Transfer only makes sense if you don't plan on colonizing. If colonization is an important part of long term planning Cycler Orbits begin to become more economical. very large Cycler Stations can solve risk factors such as radiation, micrometeorites, & crew dynamics issues.
Great video, simple and effective
very helpful. thanks.
Suppose that in a far distant future we had engines that are far more powerful than we have today; a system which negates inertia; and an endless supply of cheap power; would a Hohmann transfer still make sense in such a situation?
A Hohmann transfer is usually the most energy efficient way to get between two planets*, but it is not the most time efficient. If money, fuel and power aren't a problem, you can get to Mars much faster (on the order of a couple weeks perhaps).
A Hohmann transfer might make economic sense for large shipments which aren't needed immediately. I'm thinking along the lines of a continuous shipment of resources from the asteroid belt back to Mars/Earth. It doesn't matter that one particular rock was mined months ago if there is a continuous stream always arriving.
*Sometimes there's slightly better routes that use the gravity of multiple planets to their advantage, but these can take years longer,
Martian Colonist Thanks ^^
Is it not incorrect to divide out the mass of the spaceship? It's not a constant because it's burning fuel as it travels, thereby making it a variable mass system.
The mass is constant as you simply drift without firing your engines during the transfer (besides occasional small mid-course corrections).
For when to launch, why did you divide half our elliptic journey to mass by a full amount of mass's journey around the sun? Shouldn't it be divided by a subtraction of the journey it'll travel during our launch from earth if it's left at the r2 from r2?
Did you write it correctly and misspoke at the end? 136 degrees ahead of Earth. Right?
Great explaination, thank you!
How earth escape velocity is participating here? We can just turn off the earth gravity and orbit the sun with the V1 as the earth do.
I have a possibly stupid question: if the circular velocity of a satellite is dependent (amongst other things) upon the mass of the planet, then how can the orbital period be given only by the distance from center to satellite or semi major axis.
13:22
Can you multiply by r2² like that on the left ? Like r1² isn't multiplied at all
The orbit of the sun and mars are not perfectly round, they are elliptical as well right?
Great work! Thanks for explaining the Math :)
sir can show the values calculation of delta v1 when ever I'm the calculation I'm getting different values
Do you think that some of the colonists might be given the chance to return to Earth a few decades down the line, when the technology to do so exists?. Or would there be too many complications?
I think it will become technically possible, but the big concern is whether their metabolism, muscles and bones will even be able to cope with Earth gravity after a prolonged stay of decades in 0.38 g followed by another 8 months or so in zero g getting back. We don't know the answer to this key question at present, but I can think of a few possible ways to solve it that could allow return journeys one day.
Yeah I was thinking about that. It would be nice though if the colonists had the chance to return to Earth, almost as a retirement. So they could get a 2nd chance to explore Earth and die in a less hostile environment.
Then again, the lower gravity on Mars exerting less stress on your bones and muscles also makes for a perfect retirement ;)
Didn't think about it that way. Anyway, by the time the colonists reach retirement age, there could be quite a large colony, so there may be a lot more luxuries than the first crew would grt to enjoy.
It's certainly possible, let's not forget Elon Musk has his own plans too - "I'd like to retire to Mars some day".
So we know how to "get to Mars" as in .. orbit around the sun alongside Mars but how do you land on Mars ? I guess first slow down to start orbiting around Mars then slow down to let Mars gravity pull you in ?
could you please show the actual math calculations for delta V to make the navigation work? this way we could program the windows PC to be a real navigation computer thanks
not an expert, but I'm pretty sure that assuming the mass of the rocket stays constant during all the burns may be problematic
thanks so much this is so helpful
Another great video Ryan, keep up the good work my friend. Oh and before the RUclips grammar police arrive I will beat them to it, you missed an 'N' in the spelling of 'Transfer' on the white board at the beginning :P
Oh no, can't believe I missed that! *facepalm* (That's what happens when you do too much Maths)
Martian Colonist #quickfix
Martian Colonist Haha excellent work!!
Martian Colonist I love the subtle n at the start now
Robert Latta You didn't see anything *looks side to side* :)
Is it possible to calculate what the ratio is between the energy that is required to get to Mars, and the energy that is required to return from Mars, neglecting the changing mass of the rocket during the trip?
In this case it is easy. It is exactly the same, so the ratio you are asking for is 1.
And this is the whole point: In order to come back you need to carry all the fuel with you for the return trip. Which drastically worsens that ratio.
The two velocitys dealt with in the video do not replace the second cosmic velocity 11,2 km/s, which is needed to leave the earth orbit right?
No, these are the changes in velocity once you are already in a 'high parking orbit' above the Earth. The velocity to reach orbit or escape was derived in the previous video.
Thanks a lot :) Keep up the good work
in 15:37 where did that -1 come from ?
dV1 = Vp - V1
sustitute the "complicated" formulas for Vp and V1. Both of those parts have the sqrt(G*Ms/r1) in common (for the Vp part: you have to fiddle a little bit around), so move this part out and the -1 appears.
I personally would not have done that last step. V1 and V2 are already known numerically. So just stick with dV1 = Vp - V1, calculate Vp and subtract numerically. His final equations are i nno way "prettier" or "simpler" then just calculating Vp and Va and use those to calculate the deltas numerically. But, that is just my oppinion.
For your Vp = V1 + dV1 thought u said to subtract V1 from Vp why are u multiplying it
Awesome video! There's one thing that bothers me, though. The Earth's orbital velocity is higher then Mars's - because of being on a smaller orbit around the sun. So how is it, that in order to catch up with Mars we need to have v1 + Dv1 + Dv2, which is larger than v1 alone?
Great question!
The spacecraft's velocity when it leaves Earth orbit is v1 + Dv1, but as it travels further from the sun it's velocity decreases due to climbing further out of the Sun's gravitational potential. By the time it reaches the location of Mars' orbit, it has a new velocity, va, that is lower than v2. The second Dv2 is then required to increase va to match Mars' orbital velocity, v2.
This is summarised by the equations at the bottom at: 8:14.
+Martian Colonist Alright! So we need that Dv2 because the Sun has slowed the spacecraft down. I wonder, however, when the spacecraft gets the inital velocity of v1 + Dv1, its orbit should initally be to the 'inside' of the Eearth's orbit. That's because a larger orbital speed should mean decreasing the orbit's radius. But the way you draw it, and also the way Wikipedia draw it, the ellipse is all the time 'outside' of the Earth's orbit. Could you explain where is my thinking wrong?
Wikipedia link:
en.wikipedia.org/wiki/Hohmann_transfer_orbit#/media/File:Hohmann_transfer_orbit.svg
The issue here is the transition from a circular orbit to an elliptical orbit. v=sqrt(GM/r) is only valid for a circular orbit.
When Dv1 is added the kinetic energy of the orbiting body is increased, allowing it to move further out of the gravitational potential well of the Sun, hence the radius intuitively increases. The energy increase from Dv1 is effectively altering the orbit to include an (outwards) radial component to the velocity. An extra term must be added to the energy equation to represent this (1/2*m*(dr/dt)^2) at points in the orbit where the velocity is not entirely tangential to the radius vector (such as after the boost). Since energy and angular momentum are conserved following the boost, I chose to evaluate the mathematics at the turning points in the orbit where the equations all simplify (e.g. you can also neglect the sin(theta) term in the cross product in L=r x mv=r*m*v*sin(theta) here since theta = pi/2).
You can of course evaluate all of this by studying eccentricity changes and the geometry of the resulting ellipse, but I felt that covered up the fundamental physics of what was going on here - namely conservation of energy and angular momentum. Would have also had to introduce a lot more Maths and terminology to do it this way! :-)
So Dv1 is perpendicular to the v1 vector? And the resultant velocity vector just after the boost is not tangent to the Earth's orbit, hence the elliptical orbit. Thanks for the explanation! :)
Dv1 is actually parallel to the v1 vector, the issue is it leads to a perpendicular acceleration. Since acceleration is the rate of change of velocity, this leads to a radial velocity developing over time that turns the circular orbit into an ellipse. Hope that helps!
Wouldn't a rocket with escape velocity follow a hyperbolic orbit? Do Kepler's laws still approximate this?
If you achieve escape velocity from Earth, then yes, you will follow a hyperbolic orbit in the vicinity of Earth. However, you likely will not be at escape velocity from the Sun, so you will follow an elliptical solar orbit once outside of Earth's gravity well.
Hyperbolic orbits are also described by Kepler's first 2 laws, as these are expressions of Newtonian gravity and conservation of momentum. The third law would not hold though, as the orbital period is undefined for an unbound orbit.
@@martiancolonist I see, so it would look hyperbolic at first but quickly be redirected to a more elliptical path because of the sun's gravity? Thanks for responding, I had one tangent question; I'm in undergrad right now planned to major in CS/math but I love space travel stuff, was curious what you studied in undergrad/grad?/proposed PhD if you don't mind sharing
Plez make a video by considering the rotational speeds
I‘m wondering if finding T or theta would also be possible when the orbit of the target planet is eccentric. How would you approach something like that? Is that solvable or do you just let computers solve that problem numerically?
The time peroid can be shown to still follow Kepler's 3rd law even for an elliptical orbit (remarkably, Kepler's 3rd law even hold in General Relativity). In reality, for any system with more than 2 bodies (except under special circumstances) we have to numerically solve the equations of motion.
If you are interested in how the Maths checks out for elliptical orbits, I could dig out a paper or two?
Thanks for getting back to me this quickly. I'm playing around with Kerbal Space Program. The game makes it "easy", because one vessel is always in only one sphere of influence at a given time. It'd be great if you could dig out the papers - although I'm unsure if I'll understand them :)
Btw: I found two great website that calculate transfer windows for the bodies in the game:
#1: alexmoon.github.io/ksp
#2: ksp.olex.biz/
Is the orbit at the end orbiting around mars or is it simply following mars' path around the sun?
Mars' path. In practice, you would then have to go through orbit injection / aerobraking etc.
Thanks for Cal 3 lecture if we get there due to ai issue
Anytime.
practically every probe reaches mars in different time due to the noncircular and noncoplanar orbits of Earth and Mars ( the eccentricities of these orbits are 0.0167 and 0.0934 respectively, and the inclination of Mars to the ecliptic is 1° 51' 0".
Why u neglected the rotational speed of Earth and Mars??
Come on Ryan I'm waiting for a Mars update here.
I'm filming the next one this Sunday, targeting a Tuesday or Wednesday release.
Useful video, but if I may say that Kepler's 3rd law, requires a constant of proportionality = 4 π^2/GM , M = Mass of the Sun and G is the Gravitational constant value. You have not included in your calculation.
In the unit system where a is measured in astronomical units and P in years, the constant of proportionality is exactly 1 (by definition). In the SI system of units (which you are referring to), the constant of proportionality is actually 4 pi^2/G (M_star + M_object). Normally the mass of the orbiting object is much less than the mass of the star, which is when the approximation you quoted is valid (this does not hold for systems where the masses are similar, such as for binary stars).
Thank you for your reply, great work indeed, just posted it to my Facebook page with your permission !
Sure, glad you enjoyed the video!
Also take into account a deceleration burn when the spaceship will approach Mars so that it will remain in orbit around Mars and not escape or change direction.
Too expensive in terms of fuel, I know.
I'm not a physicist, but let ask you this.
What if we increase the deltaV1 ?
I assume that
a:) We crossing the mars orbit earlier so we don't need to travel 8 months.
b:) If we go fast enough at that point we might not need a second burst. Just a brake maneuver to fall into an orbit around Mars.
c:) This method can be the most energy efficient, but not the fastest.
If you increase deltaV1 in this configuration, you will miss Mars entirely (the transfer ellipse will extend beyond Mars). You can carry out different trajectories where you use more deltaV to get to Mars quicker, but the main difference there is that the direction of the velocity burn is different. This is substantially less efficient.
szf6111 the first burst makes the vessel intercept mars's orbit in the point shown in the video, but keep in mind that just with that burst, the vessel's orbit would end up back on earth (just follow the drawing). The second burst given at the interception point is used to enlarge the opposite side of the orbit so that it matches exactly on mars's orbit!
Increasing the DeltaV1 will just enlarge to much that part of the orbit making the vessel completely miss mars
Samuel Yepes
I get it that if we use more DeltaV1 in THIS scenario then the vessel will miss Mars.
My thinking was that assuming we can build more and more powerful boosters maybe we can cut the travel time considerably.
I just don't know how it works.
Lets say we use twice as much DeltaV1.
Should we launch when Mars is 22 degrees ahead of the Earth instead of 44 as calculated in the video?
The travel time will be also reduced to half ?
I think you would get there quicker but you would need an equal amount of fuel to save and fire retrograde halfway there (I think) to slow down enough or you would zing right by Mars
Nice Video, Thanks! It would seem that Robert Zubrin has convinced everybody that making rocket fuel (CH4 & O2) is possible from the martian atmosphere and local water. So the delta V required would be that of getting back off of the surface and then the same delta V to get back to Earth so I think the answer for the final question is more like x2.5 mass off the cuff. Has anyone done an explanation of SpaceX's BFS not needing a booster to get off of mars and back to earth?
Well. The problem is: In order to extract that fuel from mars itself, you need machinery there. Which you have to bring with you. So for the moment it is much easier to actually take the required fuel with you. And this will stay that way for a very long time.
I have not done the calculations. But Mars gravity is much weaker then earths gravity.
can someone pls tell me how he got kilometres per second because when I calculated it I got something around 1140480000
thank you!
you seem to have assumed that the rocket is already in orbit around the sun - however the majority of delta v is actually needed to get out of earth's gravity well and into solar orbit. similarly, martian gravity well is also ignored and these two account for majority of fuel burn. actual delta v for mars should be around 15-16 km/s and not 5-5.5
Correct, this video is only concerned with the Hohmann transfer itself. The delta V for escaping Earth's gravity well was covered in the previous video in this series.
just curious, whats this profession called? like if i go to a video titled "build a website", I will most likely meet people who are software engineers. In the same way, all the people here, whats this profession called?
That's tricky. Orbital Mechanics could be considered as part of Aerospace Engineering, Astronautics, and Astrophysics.
@@martiancolonist oh ok. btw if I may ask, whats your profession? or are you just doing these videos for hobby? thanks
I'm a theoretical astrophysicist researching exoplanet atmospheres. At the time I made this video, I was working on my Physics degree.
@@martiancolonist When you say theoretical astrophysicist, do you mean that you work in some university? like who funds such research? Just curious.
Yes, I work at the Carl Sagan Institute at Cornell University.
Amazing video
but it would be Fantastic if you had used these information in space travel game and landed on mars with out help from the game.
in physics the symbol "=" means that both sides are equal or what because im sure when u were calculating the delta v the energy of mars is of course not equal to earth
Yes '=' means equal. It is not the energy of the Earth of Mars, it is the total energy of the rocket when it is at the Earth is equal to the energy when it is at Mars. Conservation of energy is a fundamental principle of Physics.
Martian Colonist yea I know what it means but just to write quickly . so , in physics if I want to equate something to something like einstein when he mde his equations what should I do
I'm editing a video at the moment and then catching a flight early tomorrow. Please drop me a comment here next week, as I am very busy right now.
Martian Colonist ok im sorry , good luck
Good theoretical work, a poem for physicists.
8.5 months travel time is correct but not practical, also space travel is too dangerous. Please recalculate when Earth and Mars are in close proximity so that travel time (one way) is 3-4 months MAX.
Thanks to gravity of mars or any other planets orbiting sun, it's not neccessary to get to their speed to land. Just getting into gravitational trap is enough. It'll gently welcome us :)
great video!
but isnt the small mass in the energy equation the sum of earth AND the rocket? like E = 0,5*(me+mr)*vp² -(G*(me+mr)*M)/r1 ... Even it falls out anyway ;)
+planmaster1988 and you forgot the inclination change of about 1.8 degrees between earth and mars.. would be great having that in here too! but thanks for that good explanation!!
+planmaster1988
There's actually a much larger difference, namely that Kepler's 3rd law as formulated by Kepler isn't entirely correct. When you derive Kepler's laws from first principles (which I'll be doing in a new two part video in a few weeks), you should have M = M_sun + M_p (a correction that is not negligible in our studies of exoplanetary systems).
Yes, I neglected the inclination change for simplicity, I assumed coplanar circular orbits so I could focus on the key physical principles of conserved quantities instead of getting bogged down in pages of Maths :-)
im just waiting for someone to make a diy of this